∵cos2B+cosB+cos(A-C)=1,∴cos2B-cos(A+C)+cos(A-C)=1,即1-2sin2B-cosAcosC+sinAsinC+cosAcosC+sinAsinC=1,即sinAsinC=sin2B,由正弦定理得ac=b2,(a,b,c>0),∴a2+c2≥2ac=2b2=14.故答案为:14.